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F>=a^8/2(a^4+b^4)+b^8(b^4+c^4)+c^8/(c^4+a^4)>=(a^4+b^4+c^4)^2/4(a^4+b^4+c^4)=(a^4+b^4+c^4)/4
a^2+b^2+c^2>=ab+bc+ca=1.
3(a^4+b^4+c^4)>=(a^2+b^2+c^2)^2=1>>>a^4+b^4+c^4>=1/3
>>>F>=1/3/4=1/12
Dấu = xảy ra khi a=b=c(tự tính)
1.b)
ĐKXĐ: \(x^2+5x-2\ge0\)
PT \(\Leftrightarrow x^2+5x-2-2\sqrt{x^2+5x-2}+1=-3\)
\(\Leftrightarrow\left(\sqrt{x^2+5x-2}-1\right)^2=-3\)(vô nghiệm)
2.
\(A=\frac{1}{ab}+\frac{1}{a^2}+\frac{1}{b^2}\)\(=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{2ab}+\left(\frac{1}{a}+\frac{1}{b}\right)^2\)
Ta có: \(2ab\le\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)\(\Rightarrow\frac{1}{2ab}\ge2\)
\(\left(\frac{1}{a}+\frac{1}{b}\right)^2\ge\left(\frac{4}{a+b}\right)^2=16\)
\(\Rightarrow A\ge18\). Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
Vậy min A=18\(\Leftrightarrow a=b=\frac{1}{2}\)
Ta có:
\(A-B=\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}>0\)
Do đó: B < A và:
\(\frac{\left(a-b\right)^2}{8\left(A-B\right)}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}-\sqrt{b}\right)}{4\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)
Mà: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}=\frac{a+b+2\sqrt{ab}}{4}=\frac{a+b}{4}+\frac{\sqrt{ab}}{2}=\frac{A+B}{2}\)
\(B< A\Rightarrow B< \frac{A+B}{2}< A\left(đpcm\right)\)
3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)
áp dụng AM-GM :
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge2\sqrt{\frac{x^2y^2}{x^2y^2}}=2\)
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{xy}{yx}}=2\)
do đó \(A\ge3.2-2.8=-10\)thì \(A_{MIN}=-10\)DẤU = sảy ra khi x= y
\(Q=a+b+\frac{a^2+b^2}{a}+\frac{a^2+b^2}{b}=a+b+\frac{8}{a}+\frac{8}{b}\).
Ta dự đoán biểu thức đạt min tại \(a=b=2\) nghĩa là \(a=\frac{4}{a},b=\frac{4}{b}\) nên ta tách:
\(Q=\left(a+\frac{4}{a}\right)+\left(b+\frac{4}{b}\right)+4\left(\frac{1}{a}+\frac{1}{b}\right)\).
Áp dụng BĐT Cauchy và BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có \(Q\ge8+\frac{16}{a+b}\).
Ta lại có \(a+b\le\sqrt{2\left(a^2+b^2\right)}=4\) nên \(Q\ge12\)
Chứng minh bđt phụ: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (1)
Ta có:\(\frac{a+b}{ab}\ge\frac{4}{a+b}\)\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng với mọi \(a,b>0\))
Đặt \(A=\frac{1}{a^2+b^2}+\frac{5}{ab}+ab\)
\(\Rightarrow A=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{9}{2ab}+ab\)
Áp dụng bđt (1) ta được: \(\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}=\frac{4}{\left(a+b\right)^2}=\frac{4}{4^2}=\frac{1}{4}\)
Áp dụng bđt Cô-si với \(\frac{9}{2ab}+ab\)ta được: \(\frac{9}{2ab}+ab\ge2\sqrt{\frac{9}{2ab}.ab}=2.\sqrt{\frac{9}{2}}=\sqrt{4.\frac{9}{2}}=\sqrt{18}=3\sqrt{2}\)
\(\Rightarrow A\ge\frac{1}{4}+3\sqrt{2}\)
Vậy \(minA=3\sqrt{2}+\frac{1}{4}\)
dòng suy ra cuối cùng mình ghi lộn, phải là -8+2021 = 2013 mới đúng :v
\(a^2+\frac{8}{a^2}+\frac{b^2}{8}=8\)
\(\Leftrightarrow8a^4+64+a^2b^2=64a^2\)
\(\Leftrightarrow a^2b^2=64a^2-8a^4-64\)
\(\Leftrightarrow a^2b^2=-8\left(a^4-8a^2+8\right)\)
\(\Leftrightarrow a^2b^2=-8\left[\left(a^2-4\right)^2-8\right]\)
\(\Leftrightarrow a^2b^2=-8\left(a^2-4\right)^2+64\le64\)
\(\Leftrightarrow-8\le ab\le8\)
\(\Rightarrow A\ge-8\sqrt{6}+2021\)
dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}ab=-8\\a^2-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}b=\pm4\\a=\pm2\end{cases}}\)